Physics: Post your doubts here!

[Thought blocker]

thank you soooo sooo much bro that really helped alot! i just forogot to ask questions 12 and 15 also. sorry for all the troubles!
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_12.pdf

Both already answered
12)

15)
When a total amount of work W is done on any object, the kinetic energy of that object changes by a quantity W - if negative work is done on the object, the kinetic energy changes by a negative amount and if positive work is done on the object it's energy changes by a positive amount.

Suppose you throw an object upwards, gravity is the only force that does any work on the object. That work is negative, so the kinetic energy of the object decreases until the object comes to a stop. When it falls down, gravity does positive work on the system thus increasing it's kinetic energy.

In this case, taking the first situation, the force is constant at magnitude F, over the entire displacement s. Thus, the work done by that force is Fs. By the
Work-Kinetic Energy theorem, this is equal to the increase in kinetic energy of the system. This value is given as 4 Joules (8-4 = change in KE = +4 Joules).

In the upcoming situation, the force is 2F, displaced through a distance 2s. Thus, the total work done is 4Fs. This is also the change in kinetic energy of the object.
From above, we know that Fs is 4 Joules, so 4Fs = 4*4 = 16 Joules increase. From 4 Joules, the increase of 16 Joules takes it to 20 Joules = B.

Numeric way explanation :

When the force = F, and distance = s, work done = Fs.
Since work done = K.E. gain → Fs = 8-4 = 4J
Fs = 4JSo, when force = 2F, and distance = 2s, work done = 2F * 2s →4Fs.
Since Fs = 4J, 4Fs = 4 * 4 = 16 J.
So gain in K.E. = 16 J.
And since the kinetic energy already equals 4 J....the new K.E. = 4 + 16 → 20 J
Hence you get answer as B

 

[doremon]

Both already answered
12)

15)
When a total amount of work W is done on any object, the kinetic energy of that object changes by a quantity W - if negative work is done on the object, the kinetic energy changes by a negative amount and if positive work is done on the object it's energy changes by a positive amount.

Suppose you throw an object upwards, gravity is the only force that does any work on the object. That work is negative, so the kinetic energy of the object decreases until the object comes to a stop. When it falls down, gravity does positive work on the system thus increasing it's kinetic energy.

In this case, taking the first situation, the force is constant at magnitude F, over the entire displacement s. Thus, the work done by that force is Fs. By the
Work-Kinetic Energy theorem, this is equal to the increase in kinetic energy of the system. This value is given as 4 Joules (8-4 = change in KE = +4 Joules).

In the upcoming situation, the force is 2F, displaced through a distance 2s. Thus, the total work done is 4Fs. This is also the change in kinetic energy of the object.
From above, we know that Fs is 4 Joules, so 4Fs = 4*4 = 16 Joules increase. From 4 Joules, the increase of 16 Joules takes it to 20 Joules = B.

Numeric way explanation :

When the force = F, and distance = s, work done = Fs.
Since work done = K.E. gain → Fs = 8-4 = 4J
Fs = 4JSo, when force = 2F, and distance = 2s, work done = 2F * 2s →4Fs.
Since Fs = 4J, 4Fs = 4 * 4 = 16 J.
So gain in K.E. = 16 J.
And since the kinetic energy already equals 4 J....the new K.E. = 4 + 16 → 20 J
Hence you get answer as B

only if it asked for the change, then the answer to 15 would have been A. Thought blocker just wanted to add to the explanation don't mind.

 

[Thought blocker]

Can you also explain q8 12 13 14 15 18 21 28 31 32 thanks

8)
lets see the motion in XY
initial velocity at x=v1
final velocity at y=v2
average velocity=distance/time
(V1+V2)/2=40/12 equation 1
simplifying it we get V1+V2=6.67

now consider motion in XZ
initial velocity at x=v1
final velocity=v3
V1+V2/2=(40+40)/(12+6)
simplifying it we get V1+V3=8.89 equation 2
subtract equation 1 from 2
i mean equation 2-equation 1
V1 is cancelled and we get V3-V2=2.22
since acceleration is constant so
acceleration in yz = acceleration in xz
= change in velocity/time
=2.22/6
=0.37

12)
torque = force x perpendicular distance to centre of pivot = 200 x 0.25 = 50Nm

13)
D, Come on yaar, theory part!

14)
work done = change in g.potential energy + w.d against friction
length of slope = 1.5/sin30 = 3m
w.d against R = 3*150=450
change in g.potenial energy = 1.5*200=300
total w.d = 750J

15)
Ek = (mv^2)/2
since it is at steady speed, no net force so
Weight (mg) = retarding force (kv)
mg = kv
v = mg/ksubstitute this to Ek formula
Ek = (m(mg/f)^2)/2 = m^3 x g^2 / 2k^2

18)
Obviously C, Look, this theory questions are directly asked from book itself, so just Read 'em, I am sure, then you'll able to score good ! - 8 DAYS LEFT! HURRY UP.

Going to school. For keeping myself ready for tomorrows practical, will solve the rest later

 

[unkidd]

8)
lets see the motion in XY
initial velocity at x=v1
final velocity at y=v2
average velocity=distance/time
(V1+V2)/2=40/12 equation 1
simplifying it we get V1+V2=6.67

now consider motion in XZ
initial velocity at x=v1
final velocity=v3
V1+V2/2=(40+40)/(12+6)
simplifying it we get V1+V3=8.89 equation 2
subtract equation 1 from 2
i mean equation 2-equation 1
V1 is cancelled and we get V3-V2=2.22
since acceleration is constant so
acceleration in yz = acceleration in xz
= change in velocity/time
=2.22/6
=0.37

12)
torque = force x perpendicular distance to centre of pivot = 200 x 0.25 = 50Nm

13)
D, Come on yaar, theory part!

14)
work done = change in g.potential energy + w.d against friction
length of slope = 1.5/sin30 = 3m
w.d against R = 3*150=450
change in g.potenial energy = 1.5*200=300
total w.d = 750J

15)
Ek = (mv^2)/2
since it is at steady speed, no net force so
Weight (mg) = retarding force (kv)
mg = kv
v = mg/ksubstitute this to Ek formula
Ek = (m(mg/f)^2)/2 = m^3 x g^2 / 2k^2

18)
Obviously C, Look, this theory questions are directly asked from book itself, so just Read 'em, I am sure, then you'll able to score good ! - 8 DAYS LEFT! HURRY UP.

Going to school. For keeping myself ready for tomorrows practical, will solve the rest later

Thanks and please do tell me about the practical also i have it as well

 

[Hijab]

Thanks

 

[Hijab]

Can someone plz plz plz explain this question to me... I would really be grateful

 

[Thought blocker]

Can someone plz plz plz explain this question to me... I would really be grateful

Taking a look at the image and the details we are given, it seems to be too cumbersome to count the waves at each point and check if they are 1/8th out of phase. In fact, determining what 1/8th out of phase is would be a big headache. So, let's turn to mathematics and try to form a "phase/wave equation".
(Note - there is a real wave equation that, interestingly, applies to any wave in nature. It is a differential equation, and you do not need it for your exams, but in case you want to put this term in Google, you might not get any of the answers you are looking for)..

Okay, back to the question.

We can see for both waves that each one has a constant frequency. This is very important, because otherwise we wouldn't be able to write down a simple equation to relate the wave phases. Keeping this in mind, let's proceed.

For waveform P, we note that there are 4 wavelengths in 18 seconds. This means that 1 wavelength passes by in 18/4 = 9/2 = 4.5 seconds.
So, taking this line of reasoning further, suppose we have a time "t" in mind. How do we find out how many wavelengths have occurred in this time t?

The answer is rather simple - the number of wavelengths that have passed in time t = t/4.5 . To test this, let us put t = 18 seconds. We get, as expected, the number of wavelengths that have passed in 18 seconds = 18/4.5 = 4. Good!

Repeating this procedure for the other wave, we see that the number of wavelengths of waveform Q that pass in the same "t" seconds are t/4 wavelengths.
So when the difference between these two is 1/8 of a wavelength, then we have our answer!

We can say that the second waveform completes it's waves faster than the first waveform (it has a higher frequency) so we write

t/4 = t/4.5 + (1/8)

Rearranging, t(0.25 - 0.222) = 0.125
0.0277 * t = 0.125
t = 4.5 seconds = B.

 

[Asad Moosvi]

May someone please explain questions 13, 14 and 20?
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_12.pdf

 

[Batguy]

/

 

[Thought blocker]

Thanks and please do tell me about the practical also i have it as well

21)


28)
Simple, The flow of charge in electric field is from -ve ( tail) to +ve(head)
So Electron will repel, so would move towards the tail, i.e is downwards.
And As there wiuld be repulsion, it would repel with great force in electric field.

31) and 32) are again theory based, I won't reply to it, read chapters, and get answer yourself And still if you have doubts, don't hesitate to ask

 

[unkidd]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_11.pdf
q9 16 18 19 23 38

 

[Thought blocker]

/

We know estimate of athlete :
K.E of athlete = 4000 J

LEARN THE ESTIMATES
I have uploaded the file

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_11.pdf
q9 16 18 19 23 38

9)
Force α Acceleration
So when Force = zero, no change in speed, When force is constant, acceleration is also constant, so speed will increase linearly.

16)
The vertical components of both H and W are cancelled by the upwards vertical component of T. Furthermore, T balances out the horizontal component of H too. Thus T has to be the largest of the three. C is the only option.

18)
P = E/t
P = (1/2mv^2)/t [m = density p * volume V]
P = (1/2pVv^2)/t [Vol = Area A * length s]
P = (1/2pAlv^2)/t [l/t = speed v]
P = 1/2Alv^3

Substitute values, Answer is B.

19) sagar65265

23)
F = kx so x = F/k
For A : x = 4/k
For B : x = 3/k
For C : x = 3k
For D : x = 8/3k
So here constant k is either being multiplied or divide, hence we take k as 1 if we would have addition or subtraction included, we take k as zero
Now put 1 instead k .. A = 4, B = 3, C = 3, D = 2.5 so A is the answer

38) sagar65265

 

[Thought blocker]

May someone please explain questions 13, 14 and 20?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_12.pdf

13)
Basic concept of dynamics, forces.
B is the answer, no explanations

14)
(0.35*3*9.81)+(0.1*1.4*9.81=(0.15*6*9.81)+ans
ans=2.8Nm

20)
Take the leftmost column of liquid.

There is a rule in fluid-statics that "the pressure in a sample of liquid is the same at any vertical height, as long as the fluid at that height is part of the whole."

What that says is that in any single sample of liquid, the pressure at any height is the same. Here, it means that the pressure at the top of the leftmost column (connected to the bulb) is the same as the pressure in the next column on the right at the same depth/height. (Neat info here)

So, take the unbroken fluid column on the left. At the top of the leftmost tube (which exposes the fluid to the pressure of the bulb), the pressure at the surface is simply 16,000 Pascals, because that is the pressure in the bulb and is also the pressure with which the gas pushes on any section of the surface around it. The liquid forms part of this surface, so the pressure on the liquid is also 16,000 Pascals.

Now let's move to the next column on the right, where the liquid column is exposed to pressure P. Where it is exposed to pressure P, the surface faces a pressure of P Pascals. This is to be expected. However, as we descend down the liquid, the pressure increases by (ρgh) at a depth h, and continues decreasing with depth.

When we reach depth h1, we note that this level coincides with the vertical level of the fluid in the leftmost column. The pressure on the right side column is
P + ρgh1, while the pressure on the left hand column is 16,000 Pascals. Since these are equal, we can write

P + ρgh1 = 16,000.

Repeating that calculation on the right side, we get

P + ρgh2 = 8,000

Eliminating P,

16,000 - ρgh1 = 8,000 - ρgh2
13,600 * 9.8 * (h1-h2) = 8000
h1 - h2 = 0.06

So the difference is 0.06 meters = 6 centimeters. The only option that agrees is D.

 

[Asad Moosvi]

13)
Basic concept of dynamics, forces.
B is the answer, no explanations

14)
(0.35*3*9.81)+(0.1*1.4*9.81=(0.15*6*9.81)+ans
ans=2.8Nm

20)
Take the leftmost column of liquid.

There is a rule in fluid-statics that "the pressure in a sample of liquid is the same at any vertical height, as long as the fluid at that height is part of the whole."

What that says is that in any single sample of liquid, the pressure at any height is the same. Here, it means that the pressure at the top of the leftmost column (connected to the bulb) is the same as the pressure in the next column on the right at the same depth/height. (Neat info here)

So, take the unbroken fluid column on the left. At the top of the leftmost tube (which exposes the fluid to the pressure of the bulb), the pressure at the surface is simply 16,000 Pascals, because that is the pressure in the bulb and is also the pressure with which the gas pushes on any section of the surface around it. The liquid forms part of this surface, so the pressure on the liquid is also 16,000 Pascals.

Now let's move to the next column on the right, where the liquid column is exposed to pressure P. Where it is exposed to pressure P, the surface faces a pressure of P Pascals. This is to be expected. However, as we descend down the liquid, the pressure increases by (ρgh) at a depth h, and continues decreasing with depth.

When we reach depth h1, we note that this level coincides with the vertical level of the fluid in the leftmost column. The pressure on the right side column is
P + ρgh1, while the pressure on the left hand column is 16,000 Pascals. Since these are equal, we can write

P + ρgh1 = 16,000.

Repeating that calculation on the right side, we get

P + ρgh2 = 8,000

Eliminating P,

16,000 - ρgh1 = 8,000 - ρgh2
13,600 * 9.8 * (h1-h2) = 8000
h1 - h2 = 0.06

So the difference is 0.06 meters = 6 centimeters. The only option that agrees is D.

Hey, thanks. How about question 25 from the same paper? I don't get that.

 

[papajohn]

Q 17 Thought Blocker plzzzzzzzzzzzzzzzzzzz
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_12.pdf

 

[Thought blocker]

Hey, thanks. How about question 25 from the same paper? I don't get that.

25)
The Y is half the amplitude of X = 8/2 = 4
And In onw oscillation of Y there is the frequency of 100Hz, so what will be the frequency of Y which oscillates 3 times when X does for 1, that would be 3 * 100 = 300Hz

 

[Thought blocker]

Q 17 Thought Blocker plzzzzzzzzzzzzzzzzzzz
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_12.pdf

mgsinteta*speed=40% of total power mgsinteta=force,,,,,,force*velocity=power
(total m)(9.81)sin30)==40%
100%=ans
Hope it helps.Please tell if the answer u get by this way is differnt from mark scheme...

 

[papajohn]

mgsinteta*speed=40% of total power mgsinteta=force,,,,,,force*velocity=power
(total m)(9.81)sin30)==40%
100%=ans
Hope it helps.Please tell if the answer u get by this way is differnt from mark scheme...

Tusi great ho yar!

 

[Hijab]

Taking a look at the image and the details we are given, it seems to be too cumbersome to count the waves at each point and check if they are 1/8th out of phase. In fact, determining what 1/8th out of phase is would be a big headache. So, let's turn to mathematics and try to form a "phase/wave equation".
(Note - there is a real wave equation that, interestingly, applies to any wave in nature. It is a differential equation, and you do not need it for your exams, but in case you want to put this term in Google, you might not get any of the answers you are looking for)..

Okay, back to the question.

We can see for both waves that each one has a constant frequency. This is very important, because otherwise we wouldn't be able to write down a simple equation to relate the wave phases. Keeping this in mind, let's proceed.

For waveform P, we note that there are 4 wavelengths in 18 seconds. This means that 1 wavelength passes by in 18/4 = 9/2 = 4.5 seconds.
So, taking this line of reasoning further, suppose we have a time "t" in mind. How do we find out how many wavelengths have occurred in this time t?

The answer is rather simple - the number of wavelengths that have passed in time t = t/4.5 . To test this, let us put t = 18 seconds. We get, as expected, the number of wavelengths that have passed in 18 seconds = 18/4.5 = 4. Good!

Repeating this procedure for the other wave, we see that the number of wavelengths of waveform Q that pass in the same "t" seconds are t/4 wavelengths.
So when the difference between these two is 1/8 of a wavelength, then we have our answer!

We can say that the second waveform completes it's waves faster than the first waveform (it has a higher frequency) so we write

t/4 = t/4.5 + (1/8)

Rearranging, t(0.25 - 0.222) = 0.125
0.0277 * t = 0.125
t = 4.5 seconds = B.

Thankyou so much